Differentiable function

Suppose $h$ is bounded by one in absolute value on a punctured neighbourhood of zero. Suppose also that there are positive sequences $p_n,q_n\to 0$ with $h(p_n)=1$ and $h(q_n)=0$. Extend $F_a(x)=|x{|}^{a}h(x)$ by $F_a(0)=0$. ThenWhen the derivative exists, it equals zero. For continuity, the bound $|F_a(x)|\le |x{|}^a$ gives the epsilon-delta definition of continuity with $\delta ={\epsilon }^{1/a}$ when $a>0$. If $a\le 0$, the values $F_a(p_n)=p_n^a$ remain at least one for sufficiently large $n$, contradicting continuity at zero.

For differentiability, the difference quotient is bounded in absolute value by $|x{|}^{a-1}$, so it tends to zero when $a>1$. Conversely its values along $q_n$ are zero, forcing any possible derivative to be zero. Its values along $p_n$ equal $p_n^{a-1}$ and fail to tend to zero when $a\le 1$.

For example, $h(x)=\sin (\pi \sin (1/x))$ has the required positive sequences $p_n=1/(2\pi n+\pi /6)$ and $q_n=1/(2\pi n)$. This separates the suppression needed for continuity from the stronger suppression needed for differentiability.

Extend the displayed function by $f(0)=0$. Its difference quotient at zero is $h\sin (1/h)$, which tends to zero. Away from zero its derivative is $2x\sin (1/x)-\cos (1/x)$. At $x_k=1/(2\pi k)$ this derivative equals $-1$, whereas at $y_k=1/((2k+1)\pi )$ it equals $+1$. Thus differentiability implies continuity of the function but does not imply continuity of its derivative.

A function with $f(0)=0$ and the displayed bound has zero Fréchet derivative at zero, since $|f(x)|/|x|\to 0$. In particular, a bounded oscillatory factor multiplied by a quadratic expression is differentiable at the origin even if the oscillations have no limit themselves. For $f(x,y)=(x^2-y^2)\sin [1/(x^2-y^2)]$, extended by zero on the two diagonal lines, this damping works at the origin. At a nonzero point of either line, the quadratic factor instead vanishes to first order along a transverse direction; the difference quotient oscillates without a limit. Thus the differentiability set is the complement of those two lines together with the origin.

If $u_n\le x\le v_n$, $u_n<v_n$ and both endpoints tend to $x$, differentiability gives $g(x+h)=g(x)+g^{\prime }(x)h+r(h)$ with $r(h)=o(|h|)$. The error in the secant is bounded by $\epsilon (|v_n-x|+|u_n-x|)/(v_n-u_n)=\epsilon$ for large $n$. Thus these secants converge even when either endpoint equals $x$.

If a real function has a finite derivative at $x$, its difference quotient tends to $f^{\prime }(x)$ and is bounded for sufficiently small nonzero increments. Multiplying it by the increment shows $f(x+h)-f(x)\to 0$, proving continuity at $x$. The converse fails: the absolute value function is continuous but not differentiable at zero.

Differentiability is the existence of a derivative at a point. For maps between finite-dimensional normed vector spaces, the existence of a Fréchet derivative requires a single linear map approximating the increment with error $o(\Vert h\Vert )$. Merely having partial derivatives need not provide this approximation; continuous partial derivatives on a neighbourhood do suffice.

Set $f(0)=0$. The difference quotient has Euclidean norm $|t|$, so $f^{\prime }(0)=0$. For $t\ne 0$, differentiation gives $\Vert f^{\prime }(t){\Vert }^2=1+4t^2$. Thus this differentiable curve has a derivative discontinuous at zero, despite each scalar component satisfying the usual one-dimensional derivative properties.

The function $f(x)=x^2\sin (1/x^2)$ for $x\ne 0$, extended by $f(0)=0$, is differentiable everywhere. At zero its derivative is zero by the difference quotient, while away from zero $f^{\prime }(x)=2x\sin (1/x^2)-2x^{-1}\cos (1/x^2)$ is unbounded near zero. Differentiability of a function is therefore weaker than continuity of its derivative.

If all partial derivatives exist near a point and are continuous there, successive coordinate increments and the one-variable mean value theorem express $f(a+h)-f(a)$ as ${\sum }_j{h}_j{\partial }_{j}f(a+{\eta }_j(h))$ with ${\eta }_j(h)\to 0$. Subtracting the displayed linear map leaves $o(\Vert h\Vert )$, proving Fréchet differentiability. In two dimensions the absolute remainder is at most $ϵ(|h_1|+|h_2|)\le \sqrt{2}ϵ\Vert h\Vert$ for small enough increments.

At a point with $f(x)\ne 0$, factor the difference of squares and divide by $f(x+h)+f(x)\to 2f(x)$ to show that differentiability of $f^2$ is equivalent to that of $f$. At a zero of $f$, its square has a local minimum; if differentiable, its derivative must vanish. The difference quotient then vanishes exactly when $f(x+h{)}^2/|h|\to 0$, equivalently $f(x+h)/\sqrt{|h|}\to 0$. For example, $f(t)=|t{|}^{\alpha }$ with $1/2<\alpha <1$ is not differentiable at zero, but its square is.

For $n=1$, differentiability at $a$ means that the difference quotient has a finite two-sided limit of a function. For $n\ge 2$, require derivatives through order $n-1$ throughout some open neighbourhood of $a$, and require the final difference quotientto exist as a finite real number. The last derivative need not exist elsewhere or be continuous at $a$. In particular, existence of $f^{\prime }(a)$ by itself does not justify taking a second derivative: the function $f^{\prime }$ must first be defined in a neighbourhood of $a$.

Smoothness describes the degree of differentiability of a function. The class $C^k$ has continuous derivatives through order $k$; a smooth function is of class $C^{\infty }$.

A smooth function has continuous derivatives of every order.

A smooth function is flat at $a$ when every derivative, including order zero, vanishes at $a$. The function $e^{-1/x^2}$ off zero, extended by zero at zero, is flat there but not identically zero near zero. Thus its Taylor series does not determine it. A real analytic function that is flat at a point vanishes on a neighborhood of that point.

In any finite-dimensional Euclidean space, this extension is a smooth function and all its partial derivatives at the origin vanish. Away from zero, repeated differentiation gives finite sums of coordinate polynomials divided by powers of $|x|$, multiplied by $e^{-1/|x{|}^2}$. Every such term is bounded in absolute value by $C|x{|}^{-N}{e}^{-1/|x{|}^2}$ near zero. Since Gaussian decay dominates every power, this bound tends to zero faster than any prescribed positive power of $|x|$. Induction using difference quotients then proves smoothness and zero derivatives of every order at zero.

The origin is nevertheless a strict global minimum, because $F(x)>0$ for $x\ne 0$. All its Taylor series coefficients vanish, so the function is not real analytic at zero. Adding a constant preserves the stationary minimum but changes the zeroth-order value. An expression written only for $x\ne 0$ has no stationary point at the excluded origin until an extension is specified.

Extend this expression by zero at zero. Gaussian decay dominates every power, so the resulting function and its derivative difference quotients vanish to every order at zero. Away from zero, differentiation replaces $p(t)$ by $2t^{3}p(t)-t^2{p}^{\prime }(t)$, another polynomial. Induction therefore gives a smooth function flat at zero. If $p$ is nonzero, its Taylor series is zero while the function is nonzero at every sufficiently small nonzero argument, so it is not real analytic there.

A smooth cutoff function is a smooth function used to localize an argument, typically equal to one on one region and zero outside a slightly larger region. Such functions can be built from bump functions and smooth partitions of unity.

A bump function is a smooth function with compact support. Rescaled and translated bump functions provide local cutoffs.

A function can be differentiable even when its partial derivatives are discontinuous. For example,away from the origin, with value zero at the origin, is differentiable there while its partial derivatives oscillate along the coordinate axes.